The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Existence of model of ZF in which every uncountable cardinal is the cardinality of some power set?

Does there exist a model of ZF in which any uncountable cardinal number is equal to the cardinality of the power set of some set ? ( In ZFC it is not possible as is shown by the answers to this ...
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When do minimal subcovers always exist - without choice?

In my answer to Doubt in the definition of a compact set, I sketched a proof of the following fact: Suppose $X$ is a topological space such that every open cover of $X$ has a minimal subcover. ...
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1answer
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Tychonoff's theorem for completely regular spaces and the axiom of choice

It is well-known that Tychonoff's theorem, i.e., that the product of any set-indexed family of compact spaces is compact, is equivalent to the axiom of choice. It is also the case that if the spaces ...
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Existence in variants of the Axiom of Choice

Let $\{A_i: i \in I\}$ be a nonempty family of nonempty sets. Why is it allowed to prove the Axiom of Choice using the Well Ordering Principle as follows: There is a well-ordering of $\cup_{i \in ...
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Stochastic differential equation with weak solution for which pathwise uniqueness hold: is the axiom of choice necessary?

A stochastic differential equation with a weak solution for which pathwise uniqueness hold has a strong solution (results by Yamada and Watanabe (1971) on the weak and strong solutions). Does the ...
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1answer
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If $a\sqcup b$ and $a\times b$ biject, then $b$ either injects or surjects in-/onto $a$

Let $a$ and $b$ be sets such that there is a bijection $a\sqcup b\to a\times b$. Show, without assuming the Axiom of Choice, that there is either a surjection $b\to a$ or an injection $b\to a$. ...
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Axiom of Choice and LEFT inverse [duplicate]

I am aware of why the Axiom of Choice is equivalent to the the statement that every surjection splits. However, I don't see why we don't also need AC to show that every injection splits. In ...
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2answers
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Sum of measurable functions is measurable: countable choice required?

The standard proof that the sum of measurable functions is measurable uses countable choice, via the countable subadditivity of outer measure ($\implies$ measurable sets are closed under countable ...
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A classification of the types of poset extensions that raise local-global principle is equivelent to the axiom of choice?

EDIT: Completely overhauled the question, remove category theory from it (stated in plain set theory): The original question Adding relations to a partial order Let $X$ be a set and let $P: Pow(X) \...
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1answer
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Proving cardinality of coproduct presentation is unique without choice?

The definition of an extensive category immediately implies that given two coproduct decompositions indexed by sets of equal cardinality, if the coproduct objects are isomorphic compatibly with their ...
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1answer
48 views

Can the unit interval be the disjoint union of countably many “super-dense” parts?

I'm curious about this question in the case where $f$ is not necessarily measurable. I think what it comes down to is this: Is there an $\varepsilon<1$ and a partition of $[0,1]$ in countably ...
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1answer
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Need this transfinite construction reach a closed set in the absence of the Axiom of Choice? Or Hartogs' theorem?

In this question I'm primarily concerned with the workings of a set theory that lacks both Foundation and Choice. Given a set $X$ and a function $\sigma:\mathcal{P}(X)\to\mathcal{P}(X)$, we have a ...
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Original proof of Zorn's Lemma

On the wiki page for Zorn's Lemma it says that this lemma was Proved by Kuratowski in 1922 and independently by Zorn in 1935 but then it says: Zorn's lemma is equivalent to the well-ordering ...
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3answers
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What is an example of two sets which cannot be compared?

In set theory, if we do not assume the Axiom of Choice, we cannot prove the Trichotomy Law between cardinals. That is, we cannot prove that for any two sets, there exists an injection from one to the ...
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Why do people accept the axiom of choice given the well ordering principle?

We know without any doubt that the axiom of choice implies (in fact is equivalent to) the well ordering principle. The well ordering principle can't be true! If we take the open interval $(0,1)$ for ...
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1answer
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Non-principal ultrafilters on a set [duplicate]

Let $X$ be a set. If $X$ is finite then all ultrafilters on $X$ are principal, i.e. have the form $\{A \subseteq X : x \in A\}$ for some $x\in X$. But now suppose $X$ is infinite, say $X=\mathbb N$. ...
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1answer
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Soft Question: Why does the Axiom of Choice lead to the weirdest constructions?

I hope this is not too off-topic / soft for math.stackexchange. My basic question is: why does the Axiom of Choice allow for some of the weirdest constructions in math? I'll make a list of the weird ...
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1answer
54 views

Are there any large cardinals that are not ordinals? [closed]

In ZF, are there any useful large cardinal that cannot be well-ordered? I think that some of the partition cardinals are that way, since with AC, we cannot have $\kappa \to (\omega)^{\omega}$. Are ...
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1answer
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Axiom of Choice implies the Well-Ordering Principle

I am trying to understand the proof of this implication we were taught in my set theory module. I cannot seem to tie it together with the final line of the argument... We used this lemma: Given $F\...
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2answers
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Can every poset be toplogically sorted?

A partial ordering $\preceq'$ on $S$ is said to be compatible with another partial ordering $\preceq$ if for all $a,b,\in S$, $$ a \preceq b \Rightarrow a \preceq' b$$ Given a poset $(S, \preceq)$, ...
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1answer
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The well-ordering principle implies Zorn's Lemma

I have read and understood proofs for each implication between $AC$, $ZL$, $WO$ except this one. These proofs need about 10 lines each. Can someone share a neat, hopefully short, proof for $WO\implies ...
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1answer
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Proving the Axiom of Choice is equalivalent to the statement “If $A$ can be well-ordered, then so can $\mathcal{P}(A)$.”

I am still not completely confident in proving equivalence between the Axiom of Choice and statements such as the one posed in the title, so I want to make sure that I am on the right track. So ...
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1answer
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A puzzle concerning the axiom of choice and the reals

Recently I was told the following riddle: Let $A=(a_1,...a_n,...a_{2n},a_{2n+1})$ a 2n+1-tuple of real numbers with the following property: Whatever number $a_i$ is removed from $A$ the remaining 2n ...
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Existence of how many sets is asserted by the axiom of choice in this case?

Applying the axiom of choice to $\{\{1,2\}, \{3,4\}, \{5,6\},\ldots\}$, does only one choice set necessarily exist, or all of the $2^{\aleph_0}$ I "could have" chosen? Or something in between? It ...
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1answer
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Does the regularity of $\omega_{\alpha+1}$ need Axiom of Choice?

Many books indicate yes to this question. However, I found the only lemma they claim to use AC is the following statement: If $\{A_i\}_{i\in I}$ is a family of sets, then $|\bigcup_{i\in I}A_i|\leq|I|...
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1answer
45 views

For every cardinal $\kappa$, $\kappa^+$ is regular

Again I'm struggling with a proof from this introduction to cardinals. Lemma 2.6. For every cardinal $\kappa$, $\kappa^+$ is regular. Proof. If not, then there would be a cofinal map $f:\...
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1answer
37 views

Using the axiom of choice to choose bijections

I couldn't think of a better question title. I am trying to understand the proof of theorem 1.8 in this introduction to cardinals. Theorem 1.8. Let $\kappa\in CARD$. Let $X=\bigcup_{\alpha<\...
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1answer
41 views

Regarding Choice in fields outside set theory.

When authors say stuff like The equivalence of continuity and sequential continuity in metric spaces uses(/requires) some version of the axiom of choice. Are they assuming that we are working ...
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1answer
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Without the Axiom of Choice, $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$?

Question: In ZF (so AC does not necessarily hold) does the following claim hold? $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$ This question arose to me when reading the top ...
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1answer
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Comparability of a set and a subset of power set.

It's well known that for any set $A$, $A < P(A)$. But now, I have some question that, WITHOUT AC, can we guarantee that $A \leq X$ or $X \leq A$ whenever $X \subseteq P(A)$? Thank you.
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1answer
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Does König's inequality require axiom of choice?

König's inequality: If $\kappa_\alpha<\lambda_\alpha$ for all $\alpha<\mu$, then $\sum\kappa_\alpha<\prod\lambda_\alpha$. In order to prove this inequality, we need to provide an injection ...
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3answers
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Is the equivalence of the ascending chain condition and the maximum condition equivalent to the axiom of dependent choice?

Assuming the axiom of dependent choice, for a partially ordered set $(X,\le)$, the following statements are equivalent: $X$ fulfils the ascending chain condition, i. e. every chain $x_1\le x_2\le\...
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1answer
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Does the relationship between Jonsson Cardinals and Jonsson Algebras require the axiom of choice?

Using Skolem functions, one can see in ZFC that a cardinal $\kappa$ is Jonsson iff there are no Jonsson algebras on $\kappa$. (I.e. every algebra of size $\kappa$ has a proper subalgebra of size $\...
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Hyperplanes without Axiom of Choice

For any projective space that contains more than one point, is it possible to prove that it contains a hyperplane without using the Axiom of Choice? It's easy enough to prove that there exists a ...
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1answer
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Proving Hahn-Banach theorem directly

Is is possible to find an example of nonreflexive Banach space for which one can establish Hahn-Banach theorem directly (not referring to the axiom of choice)?
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1answer
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Hausdorff Maximal Principle

"Hausdorff's Maximal Principle" says that any partial order P has a maximal chain (chain = linear suborder). It is equivalent to the axiom of choice. If we restrict Hausdorff's Maximal Principle to ...
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1answer
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Does $\operatorname{card}(X) < \operatorname{card}(Y)$ imply $\operatorname{card}(X^2) < \operatorname{card}(Y^2)$ without choice?

I looked to see if this question was already posted, but did not find anything. Please let me know if this is a duplicate. Assume $X, Y$ are infinite sets such that there is an injection $X \to Y$ ...
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2answers
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Is the axiom of induction constructively verifiable for a non-standard model of Peano arithmetic?

There exist models of the natural numbers which include infinite numbers. Such models are called non-standard models of arithmetic. (Proof: by the compactness theorem, there exist models of Peano ...
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Why can't Axiom of Choice be proven by Rule C

Rule C is appeared in the textbook: Introduction to mathematical logic by Mendelson (Page 81 in the fourth edition). It is said "It is very common in mathematics to reason in the following way. Assume ...
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1answer
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Uncountable dense sets of reals without the axiom of choice

In the absence of AC, can there be an uncountable dense set $S\subset\mathbb R$ such that $S\cap(-\infty,a)$ is countable for each real number $a$? (Of course, since $S$ is a countable union of ...
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2answers
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Choice function without AC in special case [duplicate]

I read the Jech, Set theory, and saw following proposition. (☆) If S is a finite family of nonempty sets, existence of choice function of S can be proved without axiom of choice. I tried to prove ...
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1answer
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If every partitioning of $X$ has a choice function, is $X$ necessarily well-orderable?

Working over the ZF axioms, it's clear that if $X$ is a well-orderable set, then every partitioning of $X$ has a choice function, by choosing the minimum of each cell. Question. Does the converse ...
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1answer
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Equivalence of Taylor series and its corresponding function and Axiom for infinite summation

Given a function $f(x)$ with a taylor series expansion, is it valid to say that $$f(x)=\sum_{n=0}^{\infty}\frac{1}{n!}f^{(n)}(a)(x-a)^n$$ for all values of x irrespective of whether the taylor series ...
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Is it really necessary to choose a basis to prove that every element of a comodule is contained in a finite-dimensional subcomodule?

Let $C$ be a coalgebra over a field and $M$ be a $C$-comodule. Then it's well-known that every element of $M$ is contained in a finite-dimensional subcomodule $M' \subset M$. This is for example an ...
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Impossibility of constructing a continuum-size linearly independent set in $\Bbb R$ [duplicate]

This is a response to the following exchange at Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional? [Bill constructs a $\aleph_0$ ...
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Existence of extension of linear map and existence of subspace complement

Great answer by Asaf Karagila to my question leads me to further questions. Let say we deal with vector spaces over $\mathbb{R}.$ Here are three sentences: For every $V$ and its subspace $E\subset ...
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1answer
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Existence of vector space complement and axiom of choice

Let say we live in the category of vector spaces over $\mathbb{R}$ or $\mathbb{C}.$ Here are three sentences: Axiom of choice Every vector space has a base. For every vector space $V$ and its ...
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Does this proof of $x\in E'\rightarrow \exists \{x_n\}\subseteq E : x_n\to x$ use the axiom of choice?

Let $(X,d)$ be a metric space. Let $E$ be a subset of $X$. If $x$ is a limit point of $E$, then there exists a sequence $x_n\in E$, $n= 1,2,\dots$ such that $x_n\to x$. Proof. Pick $...
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1answer
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A set that satisfies the hypothesis of Zorn's Lemma

A set $x$ satisfies the hypothesis of Zorn's Lemma. Let $k \in x$. $\textbf{Prove:}$ There is a $z \in x$ such that $z$ is $S$-maximal in $x$ and $z=k \vee kSz $ $\textbf{Attempt:}$ ...
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When can “$j: V \rightarrow M$ is an elementary embedding” be defined in ZF?

This regards elementary embeddings of inner models of set theory. It seems that it is in general "stated" via an axiom schema each member of which states that the class function is elementary with ...